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If the greatest common factor of positive integers n and m is 15, and

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If the greatest common factor of positive integers n and m is 15, and  [#permalink]

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New post 02 Dec 2019, 00:35
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D
E

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Re: If the greatest common factor of positive integers n and m is 15, and  [#permalink]

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New post 02 Dec 2019, 17:12
1
GCD(n,m)=15
hence, we can write n=15k

\((n+x)^{32}\)

= \((15k+x)^{32}\)

=15\(k^{32}\) + 32C1 (15\(k)^{31}\)*x+.......+32C1* 15k*\((x)^{31}\)+ \(x^{32}\)

All the terms are divisible by 15 except the last one. Hence, the remainder when \((n+x)^{32}\) is divided by 15 is same as when \(x^{32}\) is divided by 15


A. 1= (15*0+1)

hence, \(1^{32}\) will give 1 as a remainder when divided by 15


B. \(4^2\)=16= (15+1)

\((4^2)^{16}\)= \((15+1)^{16}\) will give remainder 1 when divided by 15

D. 11= (15-4)
\(11^{32}\)= \((-4)^{32}\) mod 15
\(11^{32}\)= \((4)^{32}\) mod 15
\(11^{32}\)= \(1\) mod 15

E. 14= (15-1)

\((15-1)^{32}\) will give 1 as remainder when divided by 1.

C



Bunuel wrote:
If the greatest common factor of positive integers n and m is 15, and the remainder when \((n+x)^{32}\) is divided by 15 is 1, which of the following CANNOT be the value of x?

A. 1
B. 4
C. 9
D. 11
E. 14


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Re: If the greatest common factor of positive integers n and m is 15, and   [#permalink] 02 Dec 2019, 17:12
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